Question

A point source of light is kept at a distance of 15 cm from a converging lens, on its optical axis. A sharp image is formed on the screen which is placed at distance 20 cm from lens. Find the focal length of lens.

Answer 1

Given :

▪ Distance of object = 15cm

▪ Distance of image = 20cm

▪ Type of lens : converging (convex)

To Find :

▪ Focal length of lens.

SoluTioN :

☞ f is positive for converging (convex) lens.

☞ f is negative for diverging (concave) lens.

$\dashrightarrow\sf\:\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\ \\ \dashrightarrow\sf\:\dfrac{1}{20}-\dfrac{1}{(-15)}=\dfrac{1}{f}\\ \\ \dashrightarrow\sf\:\dfrac{1}{20}+\dfrac{1}{15}=\dfrac{1}{f}\\ \\ \dashrightarrow\sf\:\dfrac{3+4}{60}=\dfrac{1}{f}\\ \\ \dashrightarrow\sf\:f=\dfrac{60}{7}\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{f=8.57\:cm}}}}\:\orange{\bigstar}$

Extra Dose :

• X-coordinate of centre of curvature and focus of convex lens are negative and those for concave lens are positive.
• In case of lenses, -ve sign of v indicates virtual image and +ve sign of v indicates real image.

Answer 2

Given:-

★$\sf\ Distance \: of\:object :- u =15cm$

★$\sf\ Distance\:of\:image:- v=20cm$

To Find:-

★$\sf\ Focal\:length\:of\:lens$

Required Solution:-

$\small\underline\bold\purple{Formula\:used\:here:-}$

$\dashrightarrow\: \sf\Large\ \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\dashrightarrow\: \sf\Large\ \frac{1}{f}=\frac{1}{20} -\frac{1}{(-15)}$

$\dashrightarrow\: \sf\Large\ \frac{1}{f}=\frac{1}{20} +\frac{1}{15}$

$\dashrightarrow\: \sf\Large\ \frac{1}{f}=\frac{3\:+\:4}{60}$

$\dashrightarrow\: \sf\Large\ \frac{1}{f}=\frac{7}{60}$

$\dashrightarrow\: \sf\Large\ f=\frac{60}{7}$

$\dashrightarrow\: \underline{\boxed{\bf{\orange{f=8.5714cm}}}}$